8.5 Adding and subtracting rational expressions (Page 2/2)

Elementary algebra Page 2 / 2

$\begin{array}{l} \frac{3 x^{2} + 4 x + 5}{(x + 6) (x - 2)} + \frac{2 x^{2} + x + 6}{x^{2} + 4 x - 12} - \frac{x^{2} - 4 x - 6}{x^{2} + 4 x - 12} \\ Factor the denominators to determine if they're the same . \\ \frac{3 x^{2} + 4 x + 5}{(x + 6) (x - 2)} + \frac{2 x^{2} + x + 6}{(x + 6) (x - 2)} - \frac{x^{2} - 4 x - 6}{(x + 6) (x - 2)} \\ The denominators are the same . Combine the numerators being careful to note the negative sign . \\ \frac{3 x^{2} + 4 x + 5 + 2 x^{2} + x + 6 - (x^{2} - 4 x + 6)}{(x + 6) (x - 2)} \\ \frac{3 x^{2} + 4 x + 5 + 2 x^{2} + x + 6 - x^{2} + 4 x + 6}{(x + 6) (x - 2)} \\ \frac{4 x^{2} + 9 x + 17}{(x + 6) (x - 2)} \end{array}$

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Practice set a

Add or Subtract the following rational expressions.

$\frac{4}{9} + \frac{2}{9}$

$\frac{2}{3}$

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$\frac{3}{b} + \frac{2}{b}$

$\frac{5}{b}$

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$\frac{5 x}{2 y^{2}} - \frac{3 x}{2 y^{2}}$

$\frac{x}{y^{2}}$

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$\frac{x + y}{x - y} + \frac{2 x + 3 y}{x - y}$

$\frac{3 x + 4 y}{x - y}$

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$\frac{4 x^{2} - x + 4}{3 x + 10} - \frac{x^{2} + 2 x + 5}{3 x + 10}$

$\frac{3 x^{2} - 3 x - 1}{3 x + 10}$

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$\frac{x (x + 1)}{x (2 x + 3)} + \frac{3 x^{2} - x + 7}{2 x^{2} + 3 x}$

$\frac{4 x^{2} + 7}{x (2 x + 3)}$

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$\frac{4 x + 3}{x^{2} - x - 6} - \frac{8 x - 4}{(x + 2) (x - 3)}$

$\frac{- 4 x + 7}{(x + 2) (x - 3)}$

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$\frac{5 a^{2} + a - 4}{2 a (a - 6)} + \frac{2 a^{2} + 3 a + 4}{2 a^{2} - 12 a} + \frac{a^{2} + 2}{2 a^{2} - 12 a}$

$\frac{4 a^{2} + 2 a + 1}{a (a - 6)}$

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$\frac{8 x^{2} + x - 1}{x^{2} - 6 x + 8} + \frac{2 x^{2} + 3 x}{x^{2} - 6 x + 8} - \frac{5 x^{2} + 3 x - 4}{(x - 4) (x - 2)}$

$\frac{5 x^{2} + x + 3}{(x - 4) (x - 2)}$

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Fractions with different denominators

Sample set b

Add or Subtract the following rational expressions.

$\begin{array}{l} \frac{4 a}{3 y} + \frac{2 a}{9 y^{2}} . & The denominators are n o t the same . Find the LCD . By inspection, the LCD is 9 y^{2} . \\ \frac{}{9 y^{2}} + \frac{2 a}{9 y^{2}} & \begin{array}{l} The denominator of the first rational expression has been multiplied by 3 y, \\ so the numerator must be multiplied by 3 y . \\ 4 a \cdot 3 y = 12 a y \end{array} \\ \frac{12 a y}{9 y^{2}} + \frac{2 a}{9 y^{2}} & The denominators are now the same . Add the numerators . \\ \frac{12 a y + 2 a}{9 y^{2}} \end{array}$

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$\begin{array}{l} \frac{3 b}{b + 2} + \frac{5 b}{b - 3} . & \begin{array}{l} The denominators are n o t the same . The LCD is (b + 2) (b - 3) . \end{array} \\ \frac{}{(b + 2) (b - 3)} + \frac{}{(b + 2) (b - 3)} & \begin{array}{l} The denominator of the first rational expression has been multiplied by b - 3, \\ so the numerator must be multiplied by b - 3. 3 b (b - 3) \end{array} \\ \frac{3 b (b - 3)}{(b + 2) (b - 3)} + \frac{}{(b + 2) (b - 3)} & \begin{array}{l} The denominator of the second rational expression has been multiplied by b + 2, \\ so the numerator must be multiplied by b + 2. 5 b (b + 2) \end{array} \\ \frac{3 b (b - 3)}{(b + 2) (b - 3)} + \frac{5 b (b + 2)}{(b + 2) (b - 3)} & \begin{array}{l} The denominators are now the same . Add the numerators . \end{array} \\ \frac{3 b (b - 3) + 5 b (b + 2)}{(b - 3) (b + 2)} & = & \frac{3 b^{2} - 9 b + 5 b^{2} + 10 b}{(b - 3) (b + 2)} \\ = & \frac{8 b^{2} + b}{(b - 3) (b - 2)} \end{array}$

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$\begin{array}{l} \frac{x + 3}{x - 1} + \frac{x - 2}{4 x + 4} . & \begin{array}{l} The denominators are n o t the same . \\ Find the LCD . \end{array} \\ \frac{x + 3}{x - 1} + \frac{x - 2}{4 (x + 1)} & The LCD is (x + 1) (x - 1) \\ \frac{}{4 (x + 1) (x - 1)} + \frac{}{4 (x + 1) (x - 1)} & \begin{array}{l} The denominator of the first rational expression has been multiplied by 4 (x + 1) so \\ the numerator must be multiplied by 4 (x + 1) . 4 (x + 3) (x + 1) \end{array} \\ \frac{4 (x + 3) (x + 1)}{4 (x + 1) (x - 1)} + \frac{}{4 (x + 1) (x - 1)} & \begin{array}{l} The denominator of the second rational expression has been multiplied by x - 1 \\ so the numerator must be multiplied by x - 1. (x - 1) (x - 2) \end{array} \\ \frac{4 (x + 3) (x + 1)}{4 (x + 1) (x - 1)} + \frac{(x - 1) (x - 2)}{4 (x + 1) (x - 1)} & \begin{array}{l} The denominators are now the same. \\ Add the numerators. \end{array} \\ \frac{4 (x + 3) (x + 1) + (x - 1) (x - 2)}{4 (x + 1) (x - 1)} \\ \frac{4 (x^{2} + 4 x + 3) + x^{2} - 3 x + 2}{4 (x + 1) (x - 1)} \\ \frac{4 x^{2} + 16 x + 12 + x^{2} - 3 x + 2}{4 (x + 1) (x - 1)} & = & \frac{5 x^{2} + 13 x + 14}{4 (x + 1) (x - 1)} \end{array}$

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$\begin{array}{l} \frac{x + 5}{x^{2} - 7 x + 12} + \frac{3 x - 1}{x^{2} - 2 x - 3} & Determine the LCD . \\ \frac{x + 5}{(x - 4) (x - 3)} + \frac{3 x - 1}{(x - 3) (x + 1)} & The LCD is (x - 4) (x - 3) (x + 1) . \\ \frac{}{(x - 4) (x - 3) (x + 1)} + \frac{}{(x - 4) (x - 3) (x + 1)} & The first numerator must be multiplied by x + 1 and the second by x - 4. \\ \frac{(x + 5) (x + 1)}{(x - 4) (x - 3) (x + 1)} + \frac{(3 x - 1) (x - 4)}{(x - 4) (x - 3) (x + 1)} & The denominators are now the same . Add the numerators . \\ \frac{(x + 5) (x + 1) + (3 x - 1) (x - 4)}{(x - 4) (x - 3) (x + 1)} \\ \frac{x^{2} + 6 x + 5 + 3 x^{2} - 13 x + 4}{(x - 4) (x - 3) (x + 1)} \\ \frac{4 x^{2} - 7 x + 9}{(x - 4) (x - 3) (x + 1)} \end{array}$

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$\begin{array}{l} \frac{a + 4}{a^{2} + 5 a + 6} - \frac{a - 4}{a^{2} - 5 a - 24} & Determine the LCD. \\ \frac{a + 4}{(a + 3) (a + 2)} - \frac{a - 4}{(a + 3) (a - 8)} & The LCD is (a + 3) (a + 2) (a - 8) . \\ \frac{}{(a + 3) (a + 2) (a - 8)} - \frac{}{(a + 3) (a + 2) (a - 8)} & The first numerator must be multiplied by a - 8 and the second by a + 2. \\ \frac{(a + 4) (a - 8)}{(a + 3) (a + 2) (a - 8)} - \frac{(a - 4) (a + 2)}{(a + 3) (a + 2) (a - 8)} & The denominators are now the same . Subtract the numerators . \\ \frac{(a + 4) (a - 8) - (a - 4) (a + 2)}{(a + 3) (a + 2) (a - 8)} \\ \frac{a^{2} - 4 a - 32 - (a^{2} - 2 a - 8)}{(a + 3) (a + 2) (a - 8)} \\ \frac{a^{2} - 4 a - 32 - a^{2} + 2 a + 8}{(a + 3) (a + 2) (a - 8)} \\ \frac{- 2 a - 24}{(a + 3) (a + 2) (a - 8)} & Factor - 2 from the numerator . \\ \frac{- 2 (a + 12)}{(a + 3) (a + 2) (a - 8)} \end{array}$

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$\begin{array}{l} \frac{3 x}{7 - x} + \frac{5 x}{x - 7} . & \begin{array}{l} The denominators are n e a r l y the same. They differ only in sign. \\ Our technique is to factor - 1 from one of them. \end{array} \\ \frac{3 x}{7 - x} = \frac{3 x}{- (x - 7)} & = & \frac{- 3 x}{x - 7} & Factor - 1 from the first term . \\ \frac{3 x}{7 - x} + \frac{5 x}{x - 7} & = & \frac{- 3 x}{x - 7} + \frac{5 x}{x - 7} \\ = & \frac{- 3 x + 5 x}{x - 7} \\ = & \frac{2 x}{x - 7} \end{array}$

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Practice set b

Add or Subtract the following rational expressions.

$\frac{3 x}{4 a^{2}} + \frac{5 x}{12 a^{3}}$

$\frac{9 a x + 5 x}{12 a^{3}}$

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$\frac{5 b}{b + 1} + \frac{3 b}{b - 2}$

$\frac{8 b^{2} - 7 b}{(b + 1) (b - 2)}$

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$\frac{a - 7}{a + 2} + \frac{a - 2}{a + 3}$

$\frac{2 a^{2} - 4 a - 25}{(a + 2) (a + 3)}$

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$\frac{4 x + 1}{x + 3} - \frac{x + 5}{x - 3}$

$\frac{3 x^{2} - 19 x - 18}{(x + 3) (x - 3)}$

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$\frac{2 y - 3}{y} + \frac{3 y + 1}{y + 4}$

$\frac{5 y^{2} + 6 y - 12}{y (y + 4)}$

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$\frac{a - 7}{a^{2} - 3 a + 2} + \frac{a + 2}{a^{2} - 6 a + 8}$

$\frac{2 a^{2} - 10 a + 26}{(a - 2) (a - 1) (a - 4)}$

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$\frac{6}{b^{2} + 6 b + 9} - \frac{2}{b^{2} + 4 b + 4}$

$\frac{4 b^{2} + 12 b + 6}{{(b + 3)}^{2} {(b + 2)}^{2}}$

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$\frac{x}{x + 4} - \frac{x - 2}{3 x - 3}$

$\frac{2 x^{2} - 5 x + 8}{3 (x + 4) (x - 1)}$

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$\frac{5 x}{4 - x} + \frac{7 x}{x - 4}$

$\frac{2 x}{x - 4}$

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Sample set c

Combine the following rational expressions.

$\begin{array}{l} 3 + \frac{7}{x - 1} . & Rewrite the expression. \\ \frac{3}{1} + \frac{7}{x - 1} & The LCD is x - 1. \\ \frac{3 (x - 1)}{x - 1} + \frac{7}{x - 1} & = & \frac{3 x - 3}{x - 1} + \frac{7}{x - 1} & = & \frac{3 x - 3 + 7}{x - 1} \\ = & \frac{3 x + 4}{x - 1} \end{array}$

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$\begin{array}{l} 3 y + 4 - \frac{y^{2} - y + 3}{y - 6} . & Rewrite the expression. \\ \frac{3 y + 4}{1} - \frac{y^{2} - y + 3}{y - 6} & The LCD is y - 6. \\ \frac{(3 y + 4) (y - 6)}{y - 6} - \frac{y^{2} - y + 3}{y - 6} & = & \frac{(3 y + 4) (y - 6) - (y^{2} - y + 3)}{y - 6} \\ = & \frac{3 y^{2} - 14 y - 24 - y^{2} + y - 3}{y - 6} \\ = & \frac{2 y^{2} - 13 y - 27}{y - 6} \end{array}$

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Practice set c

Simplify $8 + \frac{3}{x - 6} .$

$\frac{8 x - 45}{x - 6}$

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Simplify $2 a - 5 - \frac{a^{2} + 2 a - 1}{a + 3} .$

$\frac{a^{2} - a - 14}{a + 3}$

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Exercises

For the following problems, add or subtract the rational expressions.

$\frac{3}{8} + \frac{1}{8}$

$\frac{1}{2}$

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$\frac{1}{9} + \frac{4}{9}$

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$\frac{7}{10} - \frac{2}{5}$

$\frac{3}{10}$

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$\frac{3}{4} - \frac{5}{12}$

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$\frac{3}{4 x} + \frac{5}{4 x}$

$\frac{2}{x}$

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$\frac{2}{7 y} + \frac{3}{7 y}$

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$\frac{6 y}{5 x} + \frac{8 y}{5 x}$

$\frac{14 y}{5 x}$

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$\frac{9 a}{7 b} + \frac{3 a}{7 b}$

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$\frac{15 n}{2 m} - \frac{6 n}{2 m}$

$\frac{9 n}{2 m}$

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$\frac{8 p}{11 q} - \frac{3 p}{11 q}$

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$\frac{y + 4}{y - 6} + \frac{y + 8}{y - 6}$

$\frac{2 y + 12}{y - 6}$

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$\frac{y - 1}{y + 4} + \frac{y + 7}{y + 4}$

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$\frac{a + 6}{a - 1} + \frac{3 a + 5}{a - 1}$

$\frac{4 a + 11}{a - 1}$

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$\frac{5 a + 1}{a + 7} + \frac{2 a - 6}{a + 7}$

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$\frac{x + 1}{5 x} + \frac{x + 3}{5 x}$

$\frac{2 x + 4}{5 x}$

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$\frac{a - 6}{a + 2} + \frac{a - 2}{a + 2}$

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$\frac{b + 1}{b - 3} + \frac{b + 2}{b - 3}$

$\frac{2 b + 3}{b - 3}$

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$\frac{a + 2}{a - 5} - \frac{a + 3}{a - 5}$

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$\frac{b + 7}{b - 6} - \frac{b - 1}{b - 6}$

$\frac{8}{b - 6}$

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$\frac{2 b + 3}{b + 1} - \frac{b - 4}{b + 1}$

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$\frac{3 y + 4}{y + 8} - \frac{2 y - 5}{y + 8}$

$\frac{y + 9}{y + 8}$

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$\frac{2 a - 7}{a - 9} + \frac{3 a + 5}{a - 9}$

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$\frac{8 x - 1}{x + 2} - \frac{15 x + 7}{x + 2}$

$\frac{- 7 x - 8}{x + 2}$

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$\frac{7}{2 x^{2}} + \frac{1}{6 x^{3}}$

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$\frac{2}{3 x} + \frac{4}{6 x^{2}}$

$\frac{2 (x + 1)}{3 x^{2}}$

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$\frac{5}{6 y^{3}} - \frac{2}{18 y^{5}}$

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$\frac{2}{5 a^{2}} - \frac{1}{10 a^{3}}$

$\frac{4 a - 1}{10 a^{3}}$

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$\frac{3}{x + 1} + \frac{5}{x - 2}$

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$\frac{4}{x - 6} + \frac{1}{x - 1}$

$\frac{5 (x - 2)}{(x - 6) (x - 1)}$

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$\frac{2 a}{a + 1} - \frac{3 a}{a + 4}$

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$\frac{6 y}{y + 4} + \frac{2 y}{y + 3}$

$\frac{2 y (4 y + 13)}{(y + 4) (y + 3)}$

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$\frac{x - 1}{x - 3} + \frac{x + 4}{x - 4}$

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$\frac{x + 2}{x - 5} + \frac{x - 1}{x + 2}$

$\frac{2 x^{2} - 2 x + 9}{(x - 5) (x + 2)}$

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$\frac{a + 3}{a - 3} - \frac{a + 2}{a - 2}$

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$\frac{y + 1}{y - 1} - \frac{y + 4}{y - 4}$

$\frac{- 6 y}{(y - 1) (y - 4)}$

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$\frac{x - 1}{(x + 2) (x - 3)} + \frac{x + 4}{x - 3}$

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$\frac{y + 2}{(y + 1) (y + 6)} + \frac{y - 2}{y + 6}$

$\frac{y^{2}}{(y + 1) (y + 6)}$

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$\frac{2 a + 1}{(a + 3) (a - 3)} - \frac{a + 2}{a + 3}$

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$\frac{3 a + 5}{(a + 4) (a - 1)} - \frac{2 a - 1}{a - 1}$

$\frac{- 2 a^{2} - 4 a + 9}{(a + 4) (a - 1)}$

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$\frac{2 x}{x^{2} - 3 x + 2} + \frac{3}{x - 2}$

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$\frac{4 a}{a^{2} - 2 a - 3} + \frac{3}{a + 1}$

$\frac{7 a - 9}{(a + 1) (a - 3)}$

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$\frac{3 y}{y^{2} - 7 y + 12} - \frac{y^{2}}{y - 3}$

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$\frac{x - 1}{x^{2} + 6 x + 8} + \frac{x + 3}{x^{2} + 2 x - 8}$

$\frac{2 (x^{2} + x + 4)}{(x + 2) (x - 2) (x + 4)}$

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$\frac{a - 4}{a^{2} + 2 a - 3} + \frac{a + 2}{a^{2} + 3 a - 4}$

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$\frac{b - 3}{b^{2} + 9 b + 20} + \frac{b + 4}{b^{2} + b - 12}$

$\frac{2 b^{2} + 3 b + 29}{(b - 3) (b + 4) (b + 5)}$

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$\frac{y - 1}{y^{2} + 4 y - 12} - \frac{y + 3}{y^{2} + 6 y - 16}$

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$\frac{x + 3}{x^{2} + 9 x + 14} - \frac{x - 5}{x^{2} - 4}$

$\frac{- x + 29}{(x - 2) (x + 2) (x + 7)}$

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$\frac{x - 1}{x^{2} - 4 x + 3} + \frac{x + 3}{x^{2} - 5 x + 6} + \frac{2 x}{x^{2} - 3 x + 2}$

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$\frac{4 x}{x^{2} + 6 x + 8} + \frac{3}{x^{2} + x - 6} + \frac{x - 1}{x^{2} + x - 12}$

$\frac{5 x^{4} - 3 x^{3} - 34 x^{2} + 34 x - 60}{(x - 2) (x + 2) (x - 3) (x + 3) (x + 4)}$

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$\frac{y + 2}{y^{2} - 1} + \frac{y - 3}{y^{2} - 3 y - 4} - \frac{y + 3}{y^{2} - 5 y + 4}$

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$\frac{a - 2}{a^{2} - 9 a + 18} + \frac{a - 2}{a^{2} - 4 a - 12} - \frac{a - 2}{a^{2} - a - 6}$

$\frac{(a + 5) (a - 2)}{(a + 2) (a - 3) (a - 6)}$

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$\frac{y - 2}{y^{2} + 6 y} + \frac{y + 4}{y^{2} + 5 y - 6}$

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$\frac{a + 1}{a^{3} + 3 a^{2}} - \frac{a + 6}{a^{2} - a}$

$\frac{- a^{3} - 8 a^{2} - 18 a - 1}{a^{2} (a + 3) (a - 1)}$

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$\frac{4}{3 b^{2} - 12 b} - \frac{2}{6 b^{2} - 6 b}$

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$\frac{3}{2 x^{5} - 4 x^{4}} + \frac{- 2}{8 x^{3} + 24 x^{2}}$

$\frac{- x^{3} + 2 x^{2} + 6 x + 18}{4 x^{4} (x - 2) (x + 3)}$

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$\frac{x + 2}{12 x^{3}} + \frac{x + 1}{4 x^{2} + 8 x - 12} - \frac{x + 3}{16 x^{2} - 32 x + 16}$

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$\frac{2 x}{x^{2} - 9} - \frac{x + 1}{4 x^{2} - 12 x} - \frac{x - 4}{8 x^{3}}$

$\frac{14 x^{4} - 9 x^{3} - 2 x^{2} + 9 x - 36}{8 x^{3} (x + 3) (x - 3)}$

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$4 + \frac{3}{x + 2}$

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$8 + \frac{2}{x + 6}$

$\frac{8 x + 50}{x + 6}$

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$1 + \frac{4}{x - 7}$

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$3 + \frac{5}{x - 6}$

$\frac{3 x - 13}{x - 6}$

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$- 2 + \frac{4 x}{x + 5}$

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$- 1 + \frac{3 a}{a - 1}$

$\frac{2 a + 1}{a - 1}$

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$6 - \frac{4 y}{y + 2}$

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$2 x + \frac{x^{2} - 4}{x + 1}$

$\frac{3 x^{2} + 2 x - 4}{x + 1}$

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$- 3 y + \frac{4 y^{2} + 2 y - 5}{y + 3}$

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$x + 2 + \frac{x^{2} + 4}{x - 1}$

$\frac{2 x^{2} + x + 2}{x - 1}$

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$b + 6 + \frac{2 b + 5}{b - 2}$

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$\frac{3 x - 1}{x - 4} - 8$

$\frac{- 5 x + 31}{x - 4}$

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$\frac{4 y + 5}{y + 1} - 9$

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$\frac{2 y^{2} + 11 y - 1}{y + 4} - 3 y$

$\frac{- (y^{2} + y + 1)}{y + 4}$

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$\frac{5 y^{2} - 2 y + 1}{y^{2} + y - 6} - 2$

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$\frac{4 a^{3} + 2 a^{2} + a - 1}{a^{2} + 11 a + 28} + 3 a$

$\frac{7 a^{3} + 35 a^{2} + 85 a - 1}{(a + 7) (a + 4)}$

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$\frac{2 x}{1 - x} + \frac{6 x}{x - 1}$

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$\frac{5 m}{6 - m} + \frac{3 m}{m - 6}$

$\frac{- 2 m}{m - 6}$

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$\frac{- a + 7}{8 - 3 a} + \frac{2 a + 1}{3 a - 8}$

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$\frac{- 2 y + 4}{4 - 5 y} - \frac{9}{5 y - 4}$

$\frac{2 y - 13}{5 y - 4}$

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$\frac{m - 1}{1 - m} - \frac{2}{m - 1}$

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Exercises for review

( [link] ) Simplify ${(x^{3} y^{2} z^{5})}^{6} {(x^{2} y z)}^{2} .$

$x^{22} y^{14} z^{32}$

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( [link] ) Write $6 a^{- 3} b^{4} c^{- 2} a^{- 1} b^{- 5} c^{3}$ so that only positive exponents appear.

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( [link] ) Construct the graph of $y = - 2 x + 4.$
An xy coordinate plane with gridlines, labeled negative five to five on both axes.

A graph of a line passing through three points with coordinates zero, four; one, two; and two, zero.

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( [link] ) Find the product: $\frac{x^{2} - 3 x - 4}{x^{2} + 6 x + 5} \cdot \frac{x^{2} + 5 x + 6}{x^{2} - 2 x - 8} .$

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( [link] ) Replace N with the proper quantity: $\frac{x + 3}{x - 5} = \frac{N}{x^{2} - 7 x + 10} .$

$(x + 3) (x - 2)$

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Solve the mean of variance

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Step 1: Find the mean. To find the mean, add up all the scores, then divide them by the number of scores. ... Step 2: Find each score's deviation from the mean. ... Step 3: Square each deviation from the mean. ... Step 4: Find the sum of squares. ... Step 5: Divide the sum of squares by n – 1 or N.

kenneth

what is error

Yakuba Reply

Is mistake done to something

Vutshila

anas

What is the life teble

anas

Jibrin

statistics is the analyzing of data

Tajudeen Reply

what is statics?

Zelalem Reply

how do you calculate mean

Gloria Reply

diveving the sum if all values

Shaynaynay

let A1,A2 and A3 events be independent,show that (A1)^c, (A2)^c and (A3)^c are independent?

Fisaye Reply

what is statistics

Akhisani Reply

data collected all over the world

Shaynaynay

construct a less than and more than table

Imad Reply

The sample of 16 students is taken. The average age in the sample was 22 years with astandard deviation of 6 years. Construct a 95% confidence interval for the age of the population.

Aschalew Reply

Bhartdarshan' is an internet-based travel agency wherein customer can see videos of the cities they plant to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400 a. what is the probability of getting more than 12,000 hits? b. what is the probability of getting fewer than 9,000 hits?

Akshay Reply

Bhartdarshan'is an internet-based travel agency wherein customer can see videos of the cities they plan to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400. a. What is the probability of getting more than 12,000 hits

Akshay

Bright

Sorry i want to learn more about this question

Bright

Someone help

Bright

a= 0.20233 b=0.3384

Sufiyan

Shaynaynay

How do I interpret level of significance?

Mohd Reply

It depends on your business problem or in Machine Learning you could use ROC- AUC cruve to decide the threshold value

Shivam

how skewness and kurtosis are used in statistics

Owen Reply

yes what is it

Taneeya

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Source: OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3

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